Life tests performed on a sample of 13 batteries of a new model indicated: (1) an average life of 75 months, and (2) a standard deviation of 5 months. Other battery models, produced by similar processes, have normally distributed life spans. The 98% confidence interval for the population mean life of the new model is . Group of answer choices

(71.28, 78.72)

Step-by-step explanation:

We have the following information from the statement:

mean (m) = 75

sample standard deviation (sd) = 5

Sample size (n) = 13

Significance level (alpha) = 1 - 0.98 = 0.02

Degrees of freedom for t-d (df) = n - 1 = 13 - 1 = 12

The critical value would be:

t (alpha / 2) / df = T (0.01) / 12 = 2,681 (this for the table)

Margin of error equals:

E = t (alpha / 2) / df * sd / n ^ (1/2), replacing:

E = 2,681 * 5/13 ^ (1/2)

E = 3.72

Therefore, the interval of 98% confidence interval would be:

75 + 3.72 = 78.72

75 - 3.72 = 71.28

(71.28, 78.72)

answer: what if i dont want to show my work boi.

step-by-step explanation: im watching dbz and eating popcorn leave meh alone

2 is the final ! : )